A short question about trivial solution and a homogenous system We have a homogenous system $Ax=0$, I read that we have a non trivial solution iff $|A|=0$, why is that ? 
So its saying that: non trivial solution $\iff x=0\iff |A|=0$ I just don't see why it's true.
If the determinant is $0$ then the matrix isn't invertible and it is LD, so if it's LD then it makes sense that it's the other way around: $x=0$ iff $|A|=0$.
 A: First of all your $A$ must be a square matrix for $|A|$ to be defined.
If $A$ is not invertible, then the columns must be LD, hence there exists a non-trivial (at least one coefficient non-zero) combination of columns which equals the zero vector. Consequently $Ax=0$ must have a non-trivial solution (because $Ax$ is the linear combination of columns of $A$).
A: Maybe this exposition will clear up some confusion. 
Let $A$ be the matrix whose columns are $v=(a,c)$ and $w=(b,d)$, so $$A=\pmatrix{a&b\cr c&d\cr}$$ and let $x=(r,s)$. Then we have $$Ax=\pmatrix{a&b\cr c&d\cr}\pmatrix{r\cr s\cr}=\pmatrix{ar+bs\cr cr+ds\cr}=r\pmatrix{a\cr c\cr}+s\pmatrix{b\cr d\cr}=rv+sw$$ which illustrates Anurag's statement that $Ax$ is a linear combination of the columns of $A$ (the coefficients in the linear combination being the entries of $x$). 
Now suppose $\det A=0$. That's $ad-bc=0$, which we can write as $a/b=c/d$ (if we assume for the time being that $b$ and $d$ are not zero). If we define $t$ by $a/b=c/d=t$, then $$A=\pmatrix{bt&b\cr dt&d\cr}$$ and if $x=(1,-t)$ then $Ax=0$. It's not hard to reverse the reasoning and go from $Ax=0$ with $x\ne0$ to the conclusion that $\det A=0$, and I encourage OP to try it. 
A: Perhaps it will help to consider the matter in terms of actual calculation.  Given an $n\times n$ matrix $A$, we consider two problems:$\def\\#1{{\bf#1}}$


*

*How do you calculate $\det A$?

*How do you find the solutions of $A\\x=\\0$, and in particular, how do you know that there is a non-trivial solution?


The short answer to both of these is: row-reduction (aka Gaussian elimination).  To calculate the determinant, we reduce $A$ to an echelon form
$$U=\pmatrix{u_{11}&u_{12}&u_{13}&\cdots&u_{1n}\cr
             0&u_{22}&u_{23}&\cdots&u_{2n}\cr
             0&0&u_{33}&\cdots&u_{3n}\cr
             \vdots&\vdots&\vdots&\ddots&\vdots\cr
             0&0&0&\cdots&u_{nn}\cr}\ ;$$
for the present we assume that we only use the operation of adding a multiple of a row to another row, and not the operations of switching rows or multiplying a row by a constant.  Then we have
$$\det A=\det U=u_{11}u_{22}u_{33}\cdots u_{nn}\ .\tag{$*$}$$
Hence, $\det A=0$ if and only if one or more of the numbers $u_{kk}$ is zero.
To solve $A\\x=\\0$ we augment $A$ with a zero vector and row-reduce in exactly the same way, giving
$$\bigl(U\mid\\0\bigr)=\pmatrix{u_{11}&u_{12}&u_{13}&\cdots&u_{1n}&|&0\cr
             0&u_{22}&u_{23}&\cdots&u_{2n}&|&0\cr
             0&0&u_{33}&\cdots&u_{3n}&|&0\cr
             \vdots&\vdots&\vdots&\ddots&\vdots&|&\vdots\cr
             0&0&0&\cdots&u_{nn}&|&0\cr}\ .$$
Now if every $u_{kk}$ is non-zero we can solve these equations successively, starting at the bottom (back-substitution) to give
$$x_n=0\,,\ x_{n-1}=0\,,\ldots,\ x_1=0\ .$$
That is, the system has only the trivial solution $\\x=\\0$.  On the other hand, suppose that one (or more) of the numbers $u_{kk}$ is zero, and concentrate on the first such $u_{kk}$.  When back-substituting we will be able to find values for $x_n,x_{n-1},\ldots,x_{k+1}$: all zero is one possibility, and there may be others too.  But now (think of the shape of $U$) there will be no equation which gives any specific value for $x_k$; we can take any value we like, and we will always be able to complete the solution by finding values for $x_{k-1},\ldots,x_1$.  What this argument shows is:

$A\\x=\\0$ has a non-trivial solution if and only if some $u_{kk}$ is zero.

However, we can see from $(*)$ that $\det A=0$ if and only if some $u_{kk}$ is zero.  Hence,

$\det A=0$ if and only if $A\\x=\\0$ has a non-trivial solution.

Final point: what about the assumption that we only used one type of row operation when finding $\det A$?  Well firstly, if you think about it a little you can see that it is always possible to reduce $A$ to echelon form by using only this operation, so the assumption does not affect the argument.  Alternatively, if we used all three types of row operations, the effect on $\det A$ would be to change sign once or more, and to multiply once or more by non-zero numbers.  So we would have
$$\det A=\pm N u_{11}u_{22}u_{33}\cdots u_{nn}\ ,$$
where $N\ne0$, and $\det A$ is still zero if and only if some $u_{kk}$ is zero.
